How to power LED with 120 volt A/C supply 1

[ShadyRaider]

I am looking for a way to power a LED from a 120 volt A/C source. I have seen some suggestions that it could be done with a resistor only and some saying a diode is needed as well. I am using a 2.2v max V-F, and 35mA max I-F LED and want a simple "power on" indicator light from the 120 volt A/C line. Do I need a diode/resistor on both the anode and cathode of the LED? I tried a suggestion I read on this site that said a 27-47k resistor would work by it self, but it pops the resistor and LED instantly. Any suggestions would be greatly appreciated
Thanks, ShadyRaider

 

[ScottyUK]

An antiparallel diode across the LED and a series capacitor is probably the lowest loss way of doing this. A resistor in series with the cap would keep any initial charging current to a safe level to avoid killing the diodes. The capacitor will drop most of the voltage without any heat loss.


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Sometimes I only open my mouth to swap feet...

 

[VEBill]

You might want to purchase an LED Night Light from your local Dollar Store and take it apart to see how they do it.

 

[nbucska]

I suggest using for 10 mA a .22 uF 600V cap in series with
a diode bridge. The LED is in the diagonal.

Alternative : .47 uF 600V is series with the antiparallel
combination of the LED and one diode.

Since the cap costs much more and is much larger,
I suggest the first.

Don't forget to add a fuse !!! It should be a
50 to 100 mA fast acting.

For diode 1N4001 would be a good choice. I would add a R=100 Ohm, 1/4W to protect the diodes in the case of short.

Here is the wire list( a=anode, c= cathode, 1,2=one or
other end)
Parts: D1 to D4, Fuse,Cap,Led,Resistor, ACline.

AC.1---R.1
R.2---F.1
F.2---C.1
C.2---D1.a---D2.c
D1.c---L.a ---D3.c
D2.a---L.c---D4.a
AC.2---D3.a---D4.c




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[BobbyNewmark]

Why not use a neon bulb and a resistor? It's hard to get wrong, and they're nigh indestructible.

 

[itsmoked]

I have not seen a neon bulb last more than 2 years. My old boss always lobbied for them and every time I acquiesced I regretted it,(via customer annoyance calls).

My power strip neons all flicker if there is ambient light on them and are OFF in the dark.

Oh oh! And the one I plugged into an outlet, without a dropping resistor, with my fingers, during an high school electronics class lecture on, "Do nothing without checking with me!", by the teacher, exploded in my fingers on the word "me". Which leads me to disagree with the indestructibility comment, though properly applied, is probably true,(if you mean the glass envelope).

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[OperaHouse]

"I read on this site that said a 27-47k resistor would work by it self, but it pops the resistor and LED instantly."

Just what was the value of that resistor? This is a good reminder that a neophyte can take an engineering tip and consider it complete engineering advice. We all tend to assume that the readers knowledge level is sufficient to know appropriate values and wattages. 27-47K can be interpreted as 27 ohms to 47K ohms, 1/8W resistors are as good as anything, and just any old capacitor can be placed across line voltage. I'm fond of capacitors, but it should be UL X2 line rated with a 220 ohm resistor to absorb some spikes. Provide adequate spacing for any component failure.

 

[ShadyRaider]

OK, First of all thanks to all for trying to help this simple "neophte" with this problem. I am lacking knowledge in this area.....that is why I am here. By having a antiparallel connection does this mean simply having the anode from LED connected to cathode of diode, and cathode from LED to anode of another diode?? Do both leads need resistance as well??, or will a resistor and series cap only be needed for one lead after antiparallel connection is made on LED with diodes?? I have some RURP3060 diodes now and some IN4007 on the way. I tried a 1/2 watt 43ohm 2% resistor on both anode and cathode swapping the hot and neutral(of the 120v AC supply) all ways possible and could not make it work. This was done with only using one resistor on one lead at a time. OperaHouse, I did not take your previous post as "complete engineering advice" I only assumed it worked!! That post said nothing of a cap across line voltage or a 220ohm resistor, only that a 27-47k resistor was all that was needed.

 

[MacGyverS2000]

A 43 ohm resistor is going to make things go boom... 4A will do that.


Dan - Owner

http://www.Hi-TecDesigns.com

 

[OperaHouse]

I think I unintentionally made my point again. My post to clarify an issue failed. The value of the resistor is intended to be between 27K and 47K.

I was called in to determine why a dozen control boards and then the replacement boards went up in smoke at a water treatment plant. These used a dropping resistor to power a couple LEDs. A 2W power resistor was burning a hole in the board and cracking. From the pieces I was able to determine the installed resistor was between 10 and 50 ohms. The correct resistance value was supposed to be 1,000 times higher. Those color codes are getting real tough to read! Factory rep swore no one else was having a problem. The correct resistor got too hot for the board anyway, I replaced it with a cap. Interesting enough, none of the LEDs ever burnt out although it did take out a couple of control transformers. There was another little 1/4W (maybe 220 ohm)that protected them and also burnt up.

 

[nbucska]

Shady:
The cap is in series, not parallel. In the circuit I suggested both the cap and the resistor are in series with
the diode bridge.

You can replace the diode bridge with an antiparallel
diode but this circuit uses only half of the current,
the other half is shorted, co you need twice larger C
for the same current.

Yes, antiparallel is oarallel but backwards ( Anode to
cathode so the arrows of the diode and LED point in the opposite direction.

The R in series with the C limits the current if the
C is shorted or when the circuit is initially connected
to the AC, to protect the diodes and give time to the fuse to responde.




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[VEBill]

The cheap made-in-China LED night lights get the job done with only 3 or 4 components.

If you want to do it 'by the book', then use an approved, frame-mounted low voltage power supply and a series resistor.

 

[ShadyRaider]

Thanks to all for the input, I finally figured it out. I connected the ground/neutral directly to the cathode of the LED and took the hot/120-ground to a 27k resistor in series to a IN4007 diode tied antiparallel,anode of led to cathode of diode, and it works fine. Thought it would need some sort of smoothing cap, but it doesn't flicker at all

 

[shockedsilly]

Of course it flickers. You don't see it while it's sitting still but wave it around a little and you'll clearly see the led flash. It's off 60 times a second for more than 2/3 the time.

I hope you used a big enough resister - probably a 1 watt minimum. I calc your circuit is about 5 mA, so the resister is dropping 120v * .005 = .6 watt. If you were using a smaller resister and running the lead at the usual brightness and current (20ma) then you'd drop 120v * .02 = 2.4 watts. That's probably a good way to fry your circuit board. Is that why you are keeping the led so dim?

Because this circuit is so wildly inefficient (the power used by the led is less than 1% of the total) I'm very interested in nbucska's post. I understand the way the cap is limiting the current, but how do I figure it for different leds and different currents. I have a Luxeon 350mA and another Luxeon K2 that's 750mA. I'm really ignorant about A/C. What rules and formulae apply? If I knew their names, I could look them up.

I also love using three more $0.03 diodes to cut the power used in half via the bridge and lower the capacitor value and cost at the same time, probably enough to save money overall. I question the fuse, though. Doesn't the resister in series with the fuse perform the same function? If you were designing it that way, would you use a normal resister or the kind that don't burn up? Could you depend upon it failing gracefully and thereby save a component or would you be risking a fire?

As my handle shows, I'm a computer guy out of his sweet spot when things are not digital.

 

[OperaHouse]

How many engineers does it take to light a LED? After all these posts, the OP ended up picking components from different posts. The anti parallel diode on the LED is for use with the capacitor. The way it is used here it only serves to make the resistor twice as hot, the diode should be in series.

The suggestion of the fuse is obvious after reading these posts. A fuse is a good idea for someone that does not have even a basic understanding of electronics. A resistor of several hundred ohms should always be used in series with a capacitor to limit inrush current. Consider LED driver ICs, the HV9923(50mA)series is one . The new crop of LED regulators are great for powering wide voltage range electronics. I have sen these used for AC-DC operation from 24V to 250V by replacing the LED with a zener.

 

[jimkirk]

For 10 mA and 120 volts (rms), the resistor value would be 120 / .01 or 12000 ohms. R = V / I

The impedance of a capacitor C is 1/j(2*pi*C*f) where the frequency "F" is in hertz. The "j" is electrical engineering for the square root of negative 1. We don't use "i" becuse we'd confuse it for current.

Plug in the numbers for an impedance of 12000 (it will be reactive, but for this type of thing the effect is the same) and you get nbucska's value 0.22 uF.

So for your 750 mA Luxeon you'll need about 15 to 20 uF. This will have to be non-polarized (no electrolytic caps) of an appropriate voltage rating, X2 rated for safety, fused, series resistor to limit the effect of line transients...The bottom line is that the cap trick works well at the night light level, but you may have difficulty scaling it up.

 

[nbucska]

For 750 mA I would suggest a cheap transformer


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[shockedsilly]

I WONDERED why the diode wasn't in series. I thought the way it was would let a low voltage small signal diode be used instead of a high voltage one, but then I looked at current prices and a 1N4007 is only $0.03 so that couldn't be it. As a kid (1960s) I was careful to not use too much diode because I couldn't afford the 1kV ones unless I really needed them. I still have a mix of them in my parts bins. That logic is gone out the window!

Thank you all for helping me understand this, and pointing out why this doesn't scale as far as I was thinking of taking it. I WAS thinking of the Luxeon LEDs as night lights equivalent to the $4 mini fluorescent ones but using much less than their 3 watts and lasting effectively forever without mercury problems at disposal. However, a 20uF 400V polyester cap is over $10 even in some quantity so it isn't the right solution. Smaller caps are cheaper, so perhaps 300mA seems to be the edge right now for this technique depending upon the balance between price, power requirements, heat, size, and quantity. I mean, one $15 cap to use with my one $10 LED isn't so outrageous but I wouldn't want to build and sell night lights engineered like this. I have more enjoyable ways to lose my money.

Now that I know to look for "impedance", I found a great tutorial at

http://www.kpsec.freeuk.com/imped.htm

that clears things up for me. JimKirk's "square root of negative 1" threw me a bit until I looked at the web site and it triggered enough memories of the plotting of Capacitive reactance and Inductive reactance.

Thanks again, everyone.

 

[nbucska]

Shady:
The dione can't be in series with the cap. The current
in the two halfperiods MUST be the same, otherwise
the DC -- resulting from the unballance -- charges up
the cap and the current stops.



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[shockedsilly]

In ShadyRaider's final circuit, the antiparallel diode/LED are in series with a resistor - there is no capacitor. That's why OperaHouse suggested the diode be in series instead.

I also took VE1BLL's suggestion and bought a cheap night light. It turns itself off when there's light, so the circuit isn't too simple, but it is obvious to see it is powered by a 474K 250T cap (.47 uF?) in parallel with a 1M 1/8 Watt resistor and in series with a 200 ohm 1/4 watt resistor, just as some on this thread suggested.

The rest of the circuit has 5 1N4007's, one small signal or zener or something, the photocell, three white LEDs in series, an SCR, and three more resistors including a 1 watt that's high temperature and mounted away from the board. It LOOKS like it turns "off" by shunting the power to the resistor so that any power savings are imaginary. I don't understand the circuit yet, so I just might be paranoid.